Question

If z satisfies $|z-1|<|z+3|$ then $\omega =2z+3-i$ satisfies

• A. $|\omega -5-i|<|\omega +3+1|$
• B. $|\omega -5|<|\omega +3|$
• C. $Im\left(i\omega \right)>1$
• D. $|arg(\omega -1)|<\frac{\pi }{2}$

Answer 1

Answer: (B), (C)

The correct options are
B $|\omega -5|<|\omega +3|$
C $Im\left(i\omega \right)>1$

$z=\frac{w-3+i}{2}$

Hence

$|z-1|<|z+3$ implies

$|\frac{w-3+i}{2}-1|<|\frac{w-3+i}{2}+3|$

$|w-3+i-2|<|w-3+i+6|$

$|w-5+i|<|w+3+i|$

But we know that

$|z+a|\le |z|+|a|$

Hence

$|w-5+i|<|w+3+i|$
$|w-5|+|-i|<|w+3|+|i|$

$|w-5|<|w+3|$

And

$iw=2iz+3i-1$

Hence

$Im\left(iw\right)>1$.